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UkoKoshka [18]
3 years ago
15

Solve this for me 15x+82= 20x +12​

Mathematics
1 answer:
Lina20 [59]3 years ago
8 0

Answer:

here

Step-by-step explanation:

15x-20x= -82+12=>

=> - 5x= -70=>

= x= 14

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We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab
777dan777 [17]

Answer:

(a) The distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of P(\bar X>0.50) is 0.50.

Step-by-step explanation:

It is provided that random variables X_{i} are independent and identically distributed Bernoulli random variables with <em>p</em> = 0.50.

The random sample selected is of size, <em>n</em> = 100.

(a)

Theorem:

Let X_{1},\ X_{2},\ X_{3},...\ X_{n} be independent Bernoulli random variables, each with parameter <em>p</em>, then the sum of of thee random variables, X=X_{1}+X_{2}+X_{3}...+X_{n} is a Binomial random variable with parameter <em>n</em> and <em>p</em>.

Thus, the distribution of X=\sum\limits^{n}_{i=1}{X_{i}} is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean <em>μ</em> and standard deviation <em>σ</em> and appropriately huge random samples (<em>n</em> > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.  

The sample size is large, i.e. <em>n</em> = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu=p=0.50

And the standard deviation of the distribution of sample mean is given by,

\sigma_{\bar x}=\sqrt{\frac{\sigma^{2}}{n}}=\sqrt{\frac{p(1-p)}{n}}=0.05

(c)

Compute the value of P(\bar X>0.50) as follows:

P(\bar X>0.50)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{0.50-0.50}{0.05})\\

                    =P(Z>0)\\=1-P(Z

*Use a <em>z</em>-table.

Thus, the value of P(\bar X>0.50) is 0.50.

8 0
3 years ago
Can anyone help me out with this answer?
Bogdan [553]

9514 1404 393

Answer:

  7×10⁻¹ +6×10⁻² +5×10⁻³

Step-by-step explanation:

Place value in any place-value number system increases by a factor of the base for each place to the left of the "decimal" point. It is reduced by a factor of the base for each place to the right of the "decimal" point.

For base-10 numbers, the powers of 10 are -1, -2, -3, ... as you go to the right of the decimal point. Hence the number 0.765 decomposes as ...

  7×10⁻¹ +6×10⁻² +5×10⁻³

3 0
2 years ago
Complete the table by squaring each positive x-value listed 2, 3,4,5,6,7,8,9,10,11
elena55 [62]

2²=4 2×2

3²=9 3×3

4²=16 4×4

5²=25 5×5

6²=36 6×6

7²=49 7×7

8²=64 8×8

9²=81 9×9

10²=100 10×10

11²=121 11×11

4 0
3 years ago
Select the number line that shows that two opposite numbers have a sum of<br> 0.
dimulka [17.4K]

Answer:

D

Step-by-step explanation:

When you have two opposite numbers on a numberline, that is, a digit that has a negative value and the same digit having a positive value, when you add them together, we would have 0.

i.e. -a + b = 0

In the above options given, the numberline that shows this particular case is the numberline in option D.

In option D numberline, the sum of -1 and +1 equals zero.

5 0
3 years ago
Read 2 more answers
The observations of 124.53, 124.55, 142.51, and 124.52 are obtained when taping the length of a line. What should the observer c
mylen [45]

Answer: the observer should consider to eliminate or to retake the third measure.

Explanation:

The four measures taken are 124.53, 124.55, 142.51 and 124.52.

As it can be easily seen, the third measure is much different from the other three. This means that something went wrong during the observation: it can be either the measure taken wrong or that the number was written wrong (if you switch the 2 and the 4 you get a number similar to the other ones).

If the third measure is not considered, an estimate of the mean would place it around 124.5, while if the outlier (the detatched number) is considered an estimate of the mean would increase to about 129.

Therefore, in order to obtain a more reliable mean, the observer should consider to eliminate or to retake the third measure.

6 0
3 years ago
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